THE LARGEST ANGLE BISECTION PROCEDURE DAN ISMAILESCU, JOEHYUN KIM, KELVIN KIM, AND JEEWOO LEE Abstract. The largest angle bisection procedure is the operation which partitions a given triangle, T , into two smaller triangles by constructing the angle bisector of the largest angle of T . Applying the procedure to each of these two triangles produces a partition of T into four smaller triangles. Continuing in this manner, after n iterations, the initial triangle is divided into 2n small triangles. We prove that as n approaches infinity, the diameters of all these 2n triangles tend to 0, the smallest angle of all these triangles is bounded away from 0, and that, with the exception of T being an isosceles right triangle, the number of dissimilar triangles is unbounded. 1. Background and motivation For a given triangle, locate the midpoint of the longest side and then connect this point to the vertex of the triangle opposite the longest side. In other words, in any given triangle draw the shortest median. This construction is known as the longest edge bisection procedure and was first considered in 1975 by Rosenberg and Stenger [3]. Let ∆01 be a given triangle. Bisect ∆01 into two triangles ∆11 and ∆12 according to the procedure defined above. Next, bisect each ∆1i, i = 1; 2, forming four new triangles ∆2i, i = n 1; 2; 3; 4. Continue in this fashion. For every nonnegative integer n set Tn = f∆ni : 1 ≤ i ≤ 2 g, n so Tn is the set of 2 triangles created in the n-th iteration. Please refer to figure 1 for an illustration of this process for n = 3. arXiv:1908.02749v2 [math.MG] 29 Sep 2019 Figure 1. The longest edge bisection procedure: the first three iterations Define mn, the mesh of Tn, to be the length of the longest side among the sides of all triangles in Tn. Similarly, let γn be the smallest angle among the angles of the triangles in Tn. Motivated by possible applications to the finite element method, Rosenberg and Stenger con- sidered the following: Problem 1.1. (a) Is it true that γn is bounded away from 0 as n ! 1? (b) Is it true that mn approaches 0 as n ! 1? S1 (c) Does the family n=0 Tn contains finitely many triangle types? Based on figure 1 it is reasonable to expect the answer to the first two questions to be affir- mative. Indeed, the first question was answered by Rosenberg and Stenger themselves. Theorem 1.2. [3] With the notations above we have that sin γ0 γn ≥ arctan ≥ γ0=2; 2 − cos γ0 where γ0 is the smallest angle of the initial triangle ∆01. Equality holds when ∆01 is an equilateral triangle. As mentioned earlier, the theorem is of interest if the mesh in the finite-element approximation of solutions of differential equations is refined in the described manner; the convergence criterion of the method is that the angles of the triangles do not tend to zero. In 1890 Schwarz [7] surprised the mathematical community by providing and explicit example of a situation in which triangles are used to approximate the area of a cylinder. In this case, the sum of the areas of the triangles may not converge to the area of the cylinder as the size of each triangle approaches zero, and the number of triangles approaches infinity, if the smallest interior angle of each triangle approaches zero. The second question was answered by Kearfott [2] a few years later. Theorem 1.3. [2] Let mn be the length of the longest side among the sides of all nth generation triangles obtained by applying the longest edge bisection procedure. Then n p !b 2 c 3 m ≤ m · and therefore m ! 0 as n ! 1: n 0 2 n p Kearfott shows that m ≤ m · ( 3=2) and then uses induction. This rate of convergence was 2 0 p successively improved by Stynes [8] and by Adler [1] who proved that m ≤ m · 3 · 2−n=2 if n p n 0 −n=2 is even and mn ≤ m0 · 2 · 2 if n is odd, with equality if the initial triangle is equilateral. Also, both Stynes' and Adler's techniques lead to an answer to the third question: the union S1 n=0 Tn contains only finitely many triangle shapes (up to similarity). 2 For a given initial triangle ∆01, it would be interesting to find a formula for the number of different similarity classes generated by the longest edge bisection procedure applied to ∆01, S1 and also an expression for the smallest N such that every triangle in n=0 Tn is similar to some SN triangle in k=0 Tk. At the time of this writing, there are several known bounds but these seem rather weak. For details the reader is referred to [5, 6]. 2. The problem and summary of results In this paper we consider a different kind of bisection procedure. Question. What if instead of bisecting the longest edge, we bisect the largest angle? For any given triangle, locate the largest angle and then construct the angle bisector of this angle - see figure 2 below. Figure 2. The first iteration of the largest angle bisection procedure. In triangle ABC we have α ≥ β ≥ γ. For each of the two newly formed triangles construct the angle bisectors of their largest angles, n and so on. As in the longest edge bisection scenario, let Tn be the set of 2 triangles obtained after the nth iteration of this operation, which we are going to call the largest angle bisection procedure. Also, let mn, the mesh of Tn, to be the length of the longest side among the sides of all triangles in Tn and let γn be the smallest angle among the angles of the triangles in Tn. 3 It is then natural to ask the same questions as in problem 1.1 for this new operation. Under the assumption of the largest angle bisection procedure we prove the following results. (1) γn = min(γ; α=2); for all n ≥ 1: (2) mn ! 0 as n ! 1; 1 [ (3) With one exception, the set Tn contains infinitely many similarity types: n=0 Notice that results (1) and (2) are similar to the ones in the original problem, while result (3) is different. The remainder of the paper is dedicated to presenting proofs of these statements. Showing (1) is very easy, and the proof of (3) is not too difficult, either. However, proving (2) is quite challenging. In fact, throughout the next three sections we build the tools needed for showing that mn ! 0. Let us start with a simple proof of (1). Theorem 2.1. Let ∆01 = ABC be an arbitrary triangle with angles α ≥ β ≥ γ. Apply the largest angle bisection procedure with ABC as the initial triangle. Then, for all n ≥ 1 we have that γn = min(γ; α=2). n th Proof. Each of the 2 triangles obtained after the n iteration has a largest angle. Let αn denote the smallest such angle. It is easy to see that (4) γn+1 ≥ min(γn; αn=2): Indeed, if γn+1 is obtained by bisecting the largest angle of some n-th generation triangle then th γn+1 ≥ αn=2. Otherwise, γn+1 appears a base angle of some n generation triangle, hence, γn+1 ≥ γn. Next we prove that (5) αn+1=2 ≥ min(γn; αn=2): th Let MNP be the n generation triangle one of whose offspring contains αn+1. Without loss of generality we can assume that αn+1 is one of the angles of triangle MQP - see figure 3. Clearly, \QMP < \MQP which implies that αn+1 = max(\MQP; \MPQ) ≥ \MQP . It follows that αn+1 ≥ \MQP = \NMQ + \MNQ = \NMP=2 + \MNQ ≥ αn=2 + γn ≥ min(αn; 2γn): Combining (4) and (5) we obtain that min(γn+1; αn+1=2) ≥ min(γn; αn=2), from which we obtain that (6) γn ≥ min(γn−1.αn−1=2) ≥ min(γn−2.αn−2=2) ≥ ::: ≥ min(γ0; α0=2) = min(γ; α=2): 4 Figure 3. αn+1 = max(\MQP; \MPQ) On the other hand, it is easy to see that for all n ≥ 1 (7) γn ≤ min(γ; α=2): Indeed, if min(γ; α=2) = γ then γ appears in some nth generation triangle for all n ≥ 0 since one ◦ never bisects angles which are less than 60 . In this case, it follows that γn ≤ γ = min(γ; α=2). Otherwise, min(γ; α=2) = α=2 then α=2 appears in some nth generation triangle for all n ≥ 1 for exactly the same reason as above. Again, we obtain that γn ≤ α=2 = min(γ; α=2). This proves inequality (7). From (6) and (7) the statement of Theorem 2.1 follows. 3. Showing that mn ! 0: Initial considerations Recall that in the longest edge bisection procedure it is relatively easy to prove that m ≤ m · p p p 2 0 b 3=2 and in general that mn+2 ≤ mn· 3=2. This eventually implies that mn ≤ m0·( 3=2) n=2c. Thus mn ! 0 exponentially and the base is an absolute constant - see figure 4 (a). Note that such a result is not possible in the largest angle bisector procedure scenario. Indeed, let ABC be a very thin isosceles triangle; then the decay of mn could be quite slow depending on the choice of ABC - see figure 4 (b). On the other hand, define the shortest altitude bisection procedure to be analogous to the longest edge bisection and the largest angle bisection operation, the only difference being that at each step we draw the altitude corresponding to the largest edge of the triangle (rather than the the median or the angle bisector) - see figure 4 (c).